Question: Solve for $x$ : $6x^2 - 66x + 168 = 0$
Solution: Dividing both sides by $6$ gives: $ x^2 {-11}x + {28} = 0 $ The coefficient on the $x$ term is $-11$ and the constant term is $28$ , so we need to find two numbers that add up to $-11$ and multiply to $28$ The two numbers $-4$ and $-7$ satisfy both conditions: $ {-4} + {-7} = {-11} $ $ {-4} \times {-7} = {28} $ $(x {-4}) (x {-7}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -4) (x -7) = 0$ $x - 4 = 0$ or $x - 7 = 0$ Thus, $x = 4$ and $x = 7$ are the solutions.